Some new results on the existence of bounded positive entire solutions for quasilinear elliptic equations

Applied Mathematics and Computation 177 (2006) 606–613 www.elsevier.com/locate/amc

Some new results on the existence of bounded positive entire solutions for quasilinear elliptic equations q Honghui Yin, Zuodong Yang

*

School of Mathematics and Computer Sciences, Nanjing Normal University, Jiangsu Nanjing 210097, PR China

Abstract In this paper, our main purpose is to establish the existence of positive bounded entire solutions of second order quasilinear elliptic equations under new conditions. The main results of the present paper are new and extend the previously known results.  2005 Elsevier Inc. All rights reserved.

1. Introduction In this paper we are concerned with the existence of positive entire solutions of second order quasilinear elliptic equations of the type p2

divðjruj

ruÞ þ f ðx; uÞ ¼ 0;

x 2 RN ;

ð1Þ

N

where f(x, u) is a continuous function on R · (0,1). By an entire solution of Eq. (1) we mean a function u 2 C1(RN) which satisfies (1) at every point of RN (see [23] and references therein). This problem appears in the study of non-Newtonian fluids [29,30] and non-Newtonian filtration [31]. The quantity p is a characteristic of the medium. Media with p > 2 are called dilatant fluids and those with p < 2 are called pseudoplastics. If p = 2, they are Newtonian fluid. When p 5 2, the problem becomes more complicated since certain nice properties inherent to the case p = 2 seem to be lose or at least difficult to verify. The main differences between p = 2 and p 5 2 can be founded in [24–26]. Equations of the type (1) have been studied by many authors when p = 2 (see, for example, [1–7,17,20] and the reference therein). To do so, the authors always use the nice properties of D, such that, maximum principle and comparison principle and so on. Meanwhile, existence and structure of solutions for such equations with p > 1 in bounded domains have also attracted much interest (see [9,11–13,18,19]). However, it seems that very q Project supported by the National Natural Science Foundation of China (no. 10571022); the Natural Science Foundation of Educational Department of Jiangsu Province (no. 04KJB110062) and the Science Foundation of Nanjing Normal University (no. 2003SXXXGQ2B37). * Corresponding author. E-mail address: zdyang _ [email protected] (Z. Yang).

0096-3003/$ - see front matter  2005 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2005.09.091

H. Yin, Z. Yang / Applied Mathematics and Computation 177 (2006) 606–613

607

a little is known about existence properties for entire solutions except for the work by [10,14,15,23]. In the case of p = 2, problem (1) has been well investigated by [1,17,33]. In [16], we obtained the existence of bounded positive entire solutions for Eq. (1). In this paper, we obtain some results under new conditions. The results obtained here are complementary and extent to the cases considered by the authors in [1,16,17,33]. We first give the following lemma. Lemma 1. Suppose that f(x, u) is defined on RN + 1 and is locally Ho¨lder continuous (with exponent k 2 (0, 1)) in N x. Suppose moreover that there exist functions v; w 2 C 1þk loc ðR Þ such that divðjrvjp2 rvÞ þ f ðx; vÞ 6 0; divðjrwj

p2

x 2 RN ;

ð2Þ N

rwÞ þ f ðx; wÞ P 0;

x2R ;

ð3Þ

and wðxÞ 6 vðxÞ;

x 2 RN ;

ð4Þ

and that f(x, u) is locally Lipschitz continuous in u on the set fðx; uÞ : x 2 RN ; wðxÞ 6 u 6 vðxÞg. Then, Eq. (1) possesses an entire solution u(x) satisfying wðxÞ 6 uðxÞ 6 vðxÞ;

x 2 RN .

Proof. Let BR be the ball with radius R in RN. Consider the boundary value problem p2

divðjruj ruÞ þ f ðx; uÞ ¼ 0; ujoBR ¼ g;

x 2 BR ;

ð5Þ ð6Þ

where g(x) is a function that satisfies w(x) 6 g 6 v(x). For (5) and (6), v(x) is still a supersolution (for every R), and w(x) is still a subsolution (for every R), and w(x) 6 v(x) in BR. By C 1;a ðBR Þ estimates in [19] and monotonic iteration [8] or [27,28], one concludes that there exists uR 2 C 1 ðBR Þ, which is a weak solution of (1) with w(x) 6 uR(x) 6 v(x) in BR. Now, we want to apply elliptic interior estimates together with a diagonal process to conclude: {uR: R P 1} has a subsequence fuRk : Rk " 1g such that fuRk g converges to a function u in RN (pointwise) and this convergence is in C1 on every compact set in RN. (Therefore, u 2 C1 and div(j$ujp2$u) + f(x, u) = 0 with w(x) 6 u(x) 6 v(x), and this concludes the proof.) Step 1. On B2, {uR: R P 2} is uniformly bounded by w(x) and v(x). Since both w(x) and v(x) are bounded functions on B2, there exists M > 0 such that kf ðx; uR ðxÞÞkL1 ðB2 Þ 6 M and kuR ðxÞkL1 ðB2 Þ 6 M, for all R P 2. From (1), uR satisfies: Z Z p jruR j ¼ fuR . B2

B2

Therefore, Z p 1=q jruR j 6 kf k0 ðmeas B2 Þ C 1 kruR kp . B2 ðp1Þ=p

. When 1 < p < N, Here 1/q + 1/p = 1, and C1 is the Sobolev embedding constant. So, kuR k1;p 6 C 2 kf k0 Np/(Np) the embedding of W 1;p (B2) implies that uR2LNp/(Np)(B2). Applying Theorem 7.1 in [21, 0 ðB2 Þ in L pp. 286–287], we obtain the estimate: supfjuR j; x 2 B2 g 6 C 3 ;

ð7Þ

here C3 = C3(kfk0). If p P N, we get (7) from the Sobolev embedding theorem. Using Theorem 1.1 in [21, p. 251] , we see that uR belongs to C a ðB2 Þ for some 0 < a < 1, and kuR kCa 6 C 4 ;

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H. Yin, Z. Yang / Applied Mathematics and Computation 177 (2006) 606–613

here C4 is determined by C3. By Proposition 3.7 in [22, p. 806] we also know that uR belongs to C 1;a ðB2 Þ and kuR kC1;a 6 C 5 . Here C5 is determined by C4. From the arguments above we see that there exists C > 0 such that kuR kC1þa ðB1 Þ 6 C

for all R P 2.

Since the embedding C1+a(B1) ! C1(B1) is compact, there exists a sequence denoted by fuR1 j gj¼1;2... (where R1j " 1), which converges in C1(B1). Let u1 ðxÞ ¼ limj!1 uR1j ðxÞ, for x 2 B1; then u1 is a solution of (5) and (6) with w(x) P u1 P v(x). Step 2. Repeat Step 1 up to the existence of the sequence fuR1j gj¼1;2... to get a subsequence fuR2m gm¼1;2... converging in C1(B2) to a limit u2. Then likewise u2 is a solution of (5) and (6) and u2 jB1 ¼ u1 . Repeat Step 1 again on B3, . . . , etc. In this way, we obtain a sequence fuRkj gj¼1;2... which converges in C1(Bk) and is a subsequence of fuRðk1Þj gj ¼ 1; 2 . . . Let uk ¼ limj!1 uRkj , then, uk is a solution of (5) and (6) in Bk and uk jBk1 ¼ uk1 . Step 3. By a diagonal process, fuRmm gm¼1;2... is a subsequence of fuRkj gj¼1;2... for every k. Thus, on Bk for each k we have lim uRmm ¼ uk .

m!1

So, if we define uðxÞ ¼ limm!1 uRmm ðxÞ, then u(x) satisfies div(j$ujp2$u) + f(x, u) = 0 and v(x) P u(x) P w(x) since v(x) P uk(x) P w(x) for every k. This completes the proof of Lemma 1. h Remark 1. A function v(x) (resp. w(x)) satisfying the differential inequality (2) (resp. (3)) is referred to as a supersolution (resp. subsolution) of (1) in RN. Lemma 2. Suppose that q(x) 2 C(RN) is nonnegative and 1 p1 Z s Z 1 1N N 1 s t wðtÞ dt ds < 1; H1 ¼ 0

ð8Þ

0

where w(r) = maxjxj=rq(x). Then the equation divðjrU jp2 rU Þ ¼ qðxÞ

has a ground state solution U in RN ;

ð9Þ

which is bounded and lim|x| ! 1U(x) = 0. Proof. Because 1 p1 Z s Z 1 1 N 1 r wðrÞ dr ds; V ðxÞ ¼ sN 1 0 jxj which is a solution for the div(j$Vjp2$V) = w(jxj) in RN and limjxj!1V(x) = 0, so V is a upper solution for (9). On the other hand, 0 is a lower solution for (9), then (9) exists a bounded entire solution. h Remark 2. If N P 3, p < N, then condition (8) of Lemma 2 is replaced by Z 1 1 1 0< rp1 wðrÞp1 dr < 1; if 1 < p 6 2; Z1 1 ðp2ÞN þ1 0< r p1 wðrÞ dr < 1; if p P 2. 1

Let J ðrÞ ¼

Z r t 0

1N

Z

s 0

1 p1

t N 1

wðsÞ ds

dt.

ðAÞ ðBÞ

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609

If fact, if 1 < p 6 2, by estimating above the integral Z t 1=ðp1Þ Z r 1N N 1 p1 t s wðsÞ ds dt. J ðrÞ 6 C 1 þ 1

0

Using the assumption N P 3 in the computation of the first integral above and Jensen’s inequality to estimate the last one, Z r Z t 3Np 1 N1 J ðrÞ 6 C 2 þ C 3 t p1 s p1 wðsÞp1 ds dt. 1

1

Computing the above integral, we obtain Z r 1 1 J ðrÞ 6 C 2 þ C 4 tp1 wðtÞp1 dt. 1

Applying (A) in the integral above we infer that H1 = limr!1J(r) < 1. On the other hand, if p P 2, set Z t H ðtÞ ¼ sN 1 wðsÞ ds 0 1

and note that either, H(t) 6 1 for t > 0 or H(t0) = 1 for some t0 > 0. In the first case, H p1 6 1, and hence, Z r Z r 1 1N 1N p1 p1 t H ðtÞ dt 6 C 5 þ t p1 dt J ðrÞ ¼ 0

1 1

so that J(r) has a finite limit because p < N. In the second case, H ðsÞp1 6 H ðsÞ for s P s0 and hence, Z s Z r 1N t p1 sN 1 wðsÞ ds dt. J ðrÞ 6 C 6 þ 1

0

Estimating and integrating by parts, we obtain Z r  Z r Z t Z r ðp2ÞNþ1 ðp2ÞN þ1 pN 1N p1 N 1 p1 p1 p1 J ðrÞ 6 C 6 þ C 7 t dt þ t wðtÞ dt  r t wðtÞ dt 6 C 8 þ C 9 t p1 wðtÞ dt. N  p 1 1 0 1 By (B), H1 = limr!1J(r) < 1. Remark 3. If q(x) = q(jxj) 2 C(RN) is nonnegative and 1 p1 Z 1 Z s 1N N 1 H1 ¼ s t qðtÞ dt ds < 1. 0

0

Then the equation p2

divðjrU j

rU Þ ¼ qðxÞU c ;

lim U ðxÞ ¼ 1

jxj!1

has an entire explosive positive radial solution if c > p  1. Remark 4. If q(x) = q(jxj) 2 C(RN) is nonnegative and 1 p1 Z s Z 1 s1N tN 1 qðtÞ dt ds < 1. H1 ¼ 0

ðCÞ

0

Then the equation p2

divðjrU j

rU Þ ¼ qðxÞU c

has an entire bounded positive radial solution if 0 < c 6 p  1. Noticeable, the condition (C) is replaced by Z 1 1=ðp1Þ ðswðsÞÞ ds < 1; 0

where w(r) = maxjxj6rq(x) (see [32]).

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2. Main results By a modification of the method given in [33], we obtain the following results. Theorem 1. Let f(x, u) verify jf(x, u)j 6 q(x)F(u) in RN · (0, 1), where q(x) 2 C(RN) satisfies Z 1 1 1 rp1 wðrÞp1 dr < 1; if 1 < p 6 2; 0< Z1 1 ðp2ÞNþ1 0< r p1 wðrÞ dr < 1; if p P 2; 1

where w(r) = maxjxj=rq(x), N P 3, N > p. Suppose moreover that F is continuous in (0, 1) satisfying lim

u!0þ

F ðuÞ ¼0 up1

or

lim

u!1

F ðuÞ ¼0 up1

ð10Þ

then Eq. (1) possesses infinitely many positive solutions in RN, and each of these solutions is also bounded from below by a positive constant. Proof. By Lemma 2, the problem (9) has a bounded solution U, since q(x) P 0, so U is nonnegative in RN. Taking any positive constant C such that V± = C ± U satisfy C1 P V+ P V P C2 > 0 in RN for some ðuÞ ðtÞ C1, C2 > 0, if limu!0þ Fup1 ¼ 0. We chose a > 0 and small enough, such that tFp1 6 C 1p 1 , for any t 6 aC1, hence " # F ðaV þ Þ p1 p2 p1 p1 divðjrðaV þ Þj rðaV þ ÞÞ þ f ðx; aV þ Þ 6 a qðxÞ þ qðxÞF ðaV þ Þ ¼ a qðxÞ 1  V 6 0. p1 þ ðaV þ Þ That means aV+ is a upper solution of (1), similarly, aV will be a subsolution of (1) for a positive and small enough. Thus by Lemma 1, there exists a solution of (1) satisfying aV+ P u P aV P aC2 > 0. Since we can choose a > 0 arbitrarily small, we see that (1) possesses infinitely many bounded positive solutions. If ðuÞ limu!1 uFp1 ¼ 0 holds, it suffices to choose a > 0 large enough and then the above arguments work. h Theorem 2. Let f(x, u) satisfies jf(x, u)j 6 q(x)F(u) on RN · (0, 1) where q(x) 2 C(RN), and F is a continuous function defined on (0, 1). There exists a positive constant e0 depending on F, such that if the problem (9) has a bounded solution U satisfying U < e0, then (1) possesses infinitely many bounded solutions in RN, which are bounded from below by a positive constant. Proof. For any r0 > 0, we define hðrÞ ¼ maxs2½r0 ;r F ðsÞ in [r0,1). Without lose of generality, we can suppose that h is positive, otherwise we have trivial positive constant solution. Thus h is positive, nondecreasing 1 ap1 Þp1 and we claim that if and satisfies F 6 h on [r0, 1). For arbitrary positive number a, we set e1 ¼ ðhðaþr 0Þ kUk1 6 e1, then v1(x) = r0 + a  ac1U(x) with c = kUk1 is a subsolution of (1). In fact p2

divðjrv1 j

rv1 Þ þ f ðx; v1 Þ P qðxÞðac1 Þ

p1

 qðxÞF ðv1 Þ ¼ qðxÞ½ðac1 Þ

p1

 F ðr0 þ a  ac1 U ðxÞ.

Thus if kUk1 6 e1, we get then  p1 a p1 ðac1 Þ P ¼ hða þ r0 Þ P hða þ r0  ac1 U ðxÞÞ P F ða þ r0  ac1 U ðxÞÞ e1 so we have p2

divðjrv1 j

rv1 Þ þ f ðx; v1 Þ P 0. p1

1

a Similarly, if we take v2(x) = r0 + a + ac1U(x) and if c ¼ kU k1 6 e2 ¼ ðhð2aþr Þp1 we obtain 0Þ

divðjrv2 jp2 rv2 Þ ¼ ðac1 Þp1 qðxÞ 6 qðxÞhð2a þ r0 Þ 6 qðxÞF ða þ r0 þ ac1 U ðxÞÞ 6 f ðx; v2 Þ; which means that v2 is a upper solution of (1). Thus when kUk1 6 e2 6 e1, since v2 P v1, Lemma 1 gives a 1 ap1 Þp1 , the solution of (1) verifying v1 6 u 6 v2. Finally, by the monotonicity of h, if we fix a and e0 ¼ ðhð2aþ2r 0Þ

H. Yin, Z. Yang / Applied Mathematics and Computation 177 (2006) 606–613

611

0

same proof works for any r 2[r0, 2r0], and for a is arbitrary, it shows that (1) possesses infinitely many bounded positive solutions if kUk1 6 e0. h Lemma 3. Suppose f is nondecreasing and satisfies Z 1 Z t dt ffiffiffiffiffiffiffiffi ffi p f ðsÞ ds. < 1; where F ðtÞ ¼ p F ðtÞ 1 0 Then

Z 1

1

ds f 1=ðp1Þ ðsÞ

ð11Þ

< 1.

ð12Þ

Proof. If we can prove that there exist positive numbers d and M such that f 1=ðp1Þ ðsÞ P dp s

ð13Þ

for s P M;

then we will be done since Z s f p=ðp1Þ ðsÞ F ðsÞ ¼ f ðtÞ dt 6 sf ðsÞ 6 dp 0

for s P M

1=p

d which, in turn, yields ðF ðsÞÞ P f 1=ðp1Þ for s P M so that (11) implies (12). To prove (13), we assume it is ðsÞ false. That is, we assume there exists an increasing sequence {sj} of real numbers such that limj!1sj = 1 and f1/(p1)(sj)/sj < 1/j for all j. Since f is increasing, we have f(s) 6 f(sj) for all s 2 [0, sj], which, in turn, produces F(s) 6 sf(s) 6 sf(sj) for s 2 [0, sj]. Hence,  ðp1Þ=p Z sj  ðp1Þ=p Z sj Z sj j p j ðp1Þ=p ðp1Þ=p 1=p 1=p 1=p ½F ðsÞ ds P ½sf ðsj Þ ds P s ds ¼ ðsj  s1 Þ s p  1 s j j s1 s1 s1 p ðp1Þ=p ðp1Þ=p ðjÞ ¼ ð1  ðs1 =sj Þ Þ!1 p1

as j ! 1, contradicting (11). Thus (13) must be true. This completes the proof.

h

Theorem 3. Let f(x, u) satisfy 0 6 f(x, u) 6 q(x)F(u) on RN · (0, 1) for some positive nondecreasing and continuous function F defined on (0, 1), with F satisfies (11) and q 2 C(RN). Suppose moreover that the solution of (9) exists and verifies Z 1 dt kU k1 < ð14Þ 1 . 0 ðF ðtÞÞp1 Then the equation divðjrujp2 ruÞ ¼ f ðx; uÞ

in RN

ð15Þ N

possesses infinitely many bounded positive solutions in R , which are bounded from below by a positive constant. Proof. By hypothesis and Lemma 2, there are positive numbers a < b < 1 such that 0 6 U < we define then a function v with values in [a, b], such that Z b dt U ðxÞ ¼ 1 . vðxÞ ðF ðtÞÞp1 Clearly, v is well defined. Then a direct computation shows that p2

qðxÞ ¼ divðjrU jp2 rU Þ ¼ 

p2

divðjrvj rvÞ F 0 ðvÞ divðjrvj rvÞ þ 2 jrvjp P F ðvÞ F ðvÞ F ðvÞ

Rb a

dt 1

F ðtÞp1

<1

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H. Yin, Z. Yang / Applied Mathematics and Computation 177 (2006) 606–613

in RN, according to the fact that F 0 P 0. This means that p2

divðjrvj

rvÞ P qðxÞF ðvÞ P f ðx; vÞ

so v(x) is a subsolution of (15). As w = b is obvious a super-solution and w P v, we obtain a bounded positive solution u of (15) with w P u P v P a > 0. Using v 6 u 6 b, since b can be chosen as any positive constant large enough. We claim that there exist infinitely many bounded positive entire solutions, which proves the theorem. h Theorem 4. Let f(x, u) satisfy 0 6 f(x, u) 6 q(x)F(u) on RN · (0, 1) for some positive nondecreasing and continuous function F defined on (0, 1), and q 2 C(RN). Suppose moreover that the solution of (9) exists and verifies 1  p1 p1 t . kU k1 < sup F ðtÞ Then Eq. (1) possesses infinitely many bounded positive solutions in RN, which are bounded from below by a positive constant. p1

1

Proof. We use the same strategy as in the proof of Theorem 3. Let a > 0 be chosen such that kU k < ðaF ðaÞ Þp1 , 1 ap1 p1 using the continuity of function f there is some e > 0, such that 0 6 U 6 ðF ðaþeÞ Þ , we note first that v1  e is a subsolution. Now let v2(x) = ac1U(x) + e where c ¼ supRN U . Hence divðjrv2 jp2 rv2 Þ ¼ ðac1 Þp1 qðxÞ P qðxÞf ða þ eÞ P qðxÞF ðv2 Þ P f ðx; v2 Þ in RN, so v2 is a super-solution of (1). Therefore there exists a solution u of (1) such that v2 P u P e. As e can be chosen arbitrary small, we get infinitely many bounded positive entire solutions of (1). h Example 1. Consider the equation p2

divðjruj

g

ruÞ þ cðxÞuc ½lnð1 þ uÞ ¼ 0;

ð16Þ

where c(x) is locally Ho¨lder continuous (with exponent k 2 (0, 1)) in RN, and c and g are constants. Let ðuÞ F(u) = uc[ln(1 + u)]g. It is easy to check that (i) F(u) satisfies condition limu!0 uFp1 ¼ 0, if c > p  1 and F ðuÞ g P 0 or if c = p  1 and g > 0; (ii) F(u) satisfies limu!1 up1 ¼ 0, if c = p  1 and g < 0, 0 < c < p  1 and g is arbitrary. Suppose that there is a locally Holder continuous(with exponent k) function w(r) such that jc(x)j 6 w(jxj) for x 2 RN and condition (8) holds. Then from Theorem 2 it follows that in each of the cases mentioned above Eq. (16) processes infinitely many positive entire solutions which are bounded in RN. Example 2. Consider the equation divðjrujp2 ruÞ þ cðxÞð1  cos uÞup2 ¼ 0;

ð17Þ ðuÞ limu!0 uFp1

¼ 0, and so if c(x) is as in where p P 2. The function F ðuÞ ¼ up2 ð1  cos uÞ satisfies condition Example 1, then Eq. (17) has infinitely many solutions u(x) which are bounded in RN. References [1] T. Kusano, S. Oharu, Bounded entire solutions of second order semilinear elliptic equations with application to a parabolic initial value problem, Indiana Univ. Math. J. 34 (1) (1985) 85–95. [2] H. Berestycki, P.L. Lions, Nonlinear scalar field equations, I,II, Arch, Rational Mech. Anal. 82 (1983) 313–345, 347–375. [3] H. Berestycki, P.L. Lions, L.A. Peletier, An O.D.E. approach to the existence of positive solutions for semilinear problems in Rn, Indiana Univ. Math. J. 30 (1981) 141–157. [4] N. Kawano, On bounded entire solutions of semilinear elliptic equations, Hiroshima Math. J. 13 (1983) 125–158. [5] W.-M. Ni, On the elliptic equation Du + K(x)u(n+2)/(n  2) = 0, its generalizations, and applications in geometry, Indiana Univ. Math. J. 31 (1982) 493–529. [6] W.M. Ni, On the elliptic equation Du + K(x)e2u = 0 and conformal metrics with prescribed Gaussian curvatures, Invent. Math. 66 (1982) 343–352. [7] W. Walter, Entire solutions of the differential equation Du = f(u), J. Aust. Math. Soc. Ser. A 30 (1981) 366–368.

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